Stoichiometry

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Stoichiometry Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet “In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”

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Stoichiometry. “In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”. Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet. Stoichiometry. - PowerPoint PPT Presentation

Transcript of Stoichiometry

Page 1: Stoichiometry

Stoichiometry

Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”

Page 2: Stoichiometry

Stoichiometry

Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

(Stoy-key-aw-met-tree)

Page 3: Stoichiometry

Stoichiometry

Chemists spend a lot of time and energy making things. Just like a chef spends lots of time and effort making food, a chemist spends lots of time making chemicals. Chefs use recipes to make food, and chemists use recipes to make chemicals.

(Stoy-key-aw-met-tree)

Page 4: Stoichiometry

Stoichiometry• This my recipe for one batch of home-made

brownies to be cooked in a 9 in x 13 in pan.

2 sticks margarine• 4 eggs• 1.67 cups sugar (which is of course 1 & 2/3 cups)• 1.5 cups flour• 1 tsp salt• 2 tsp vanilla• 4 squares unsweetened choc• ¼ tsp baking powder• 1 ¾ tsp baking soda• 1 tsp salt

2M + 4E + 1.67 S + 1.5F (+ more) = 1 batch

Page 5: Stoichiometry

The Balanced Equation is a Recipe

If you had 8 eggs, (and everything else you need) how many batches of brownies could you make??

2M + 4E + 1.67 S + 1.5F (+ more) = 1 batch

8 egg * 1 batch/4eggs = 2 batches

Using the 8 eggs, how many sticks of margarine would you need to take out of the fridge?

8 eggs * 2marg/4eggs = 4 sticks margarine

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The Balanced Equation is a Recipe

If you had 4.5 cups of flour and wanted to make as many brownies as possible, how many eggs would you need to go with your 4.5 cups of flour

2M + 4E + 1.67 S + 1.5F (+ more) = 1 batch

4.5 cups flour * 4eggs/1.5cups flour = 12 eggs

What if you had 3 eggs, and you were really hungry and wanted to make the most brownies that you could, how many cups of sugar would you need to go with those eggs? 3 eggs * 1.67 cups sugar/4 eggs = 1.25 cups sugar

Page 7: Stoichiometry

The Balanced Equation is a Recipe

What if you wanted to make 3 batches of brownies, how many cups of flour (and of course all the other necessary ingredients) would you need? Use the flour to batch ratio to calculate.

2M + 4E + 1.67 S + 1.5F (+ more) = 1 batch

3 batches * 1.5 cups flour/1 batch = 4.5 cups flour

What if you had 2.5 cups of sugar and you wanted to know how much flour you would need to go with that sugar if you wanted to use it all up?

2.5 cups sugar * 1.5cups flour/1.67 cup sugar = 2.25 cups flour

Page 8: Stoichiometry

The Balanced Equation is a Recipe

You see, deciding how much of each ingredient, how many pans you need (or batches you can make) is nothing more than manipulating the ratios or proportions that exist between the ingredients. If you didn't maintain those set proportions, it is likely that your brownies would not taste very good.

Page 9: Stoichiometry

Working a Stoichiometry Problem

Write the balanced equation that represents the synthesis of ammonia from nitrogen gas and hydrogen gas

N2 + 3 H2 → 2 NH3

Given 7 moles of nitrogen gas, how many moles of hydrogen gas do you need to go with it?

7 moles N2 * 3H2/1N2 = 21 moles H2

Page 10: Stoichiometry

Working a Stoichiometry Problem

Given 7 moles of nitrogen gas, how many moles of ammonia can you make?

7 moles N2 * 2NH3/1N2 = 14 moles NH3 can be produced

Would your answer be the same if you started with 21 moles of hydrogen?

and of course its the same even when starting with 21 moles H2 * 2NH3/3N2 = = 14 moles NH3 can be produced

N2 + 3 H2 → 2 NH3

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Using mass measurements in the problem

• Since in the lab we have no method to directly measure moles, we usually measure quantities in mass.

• Before using the balanced equation (mole ratios – stoichiometric link) to do any problems, you must first change any information given in grams to moles.

19.6 g N2 * (1mol/28.02g) * (3H2/1N2) * (2.02g/1mol) =

Given 19.6 g of nitrogen gas, how many grams of hydrogen gas would you need to go with it?

N2 + 3 H2 → 2 NH3

4.24 g of H2 needed to go with the nitrogen

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Using mass measurements in the problem

19.6 g N2 * (1mol/28.02g) * (2NH3/1N2) * (17.04g/1mol) =

Given 19.6 g of nitrogen gas, determine the mass of ammonia that you can make.

N2 + 3 H2 → 2 NH3

23.8 g of NH3 can be produced,

and of course its the same even when starting with 4.24 g of H2* (1mol/2g) * (2NH3/3H2)* (17.04g/1mol) =

23.8 g of NH3 can be

produced,

Would your answer be the same if you had started with 4.24 g of hydrogen (the amount of gas that would also be required to do this reaction.)

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Understanding all of the steps

Write a balanced equation that represents the reaction between magnesium and aluminum sulfate.

8.4 g * 1mol Mg * 1Al2(SO4)3 * 342.17 g = 24.3g 3Mg 1 mol Al2(SO4)3

b) From this combination in part a, what mass of aluminum could you produce?

8.4 g * 1mol/24.3g * 2Al/3Mg * 27g/1mol =

3 Mg + Al2(SO4)3 → 3 MgSO4 + 2 Ala) If you had 8.40 g of magnesium, what mass of aluminum sulfate would you need to completely react with it.

c) Suppose you actually did this experiment with the quantities described above, and you were able to produce 5.10 g of aluminum. Calculate the % yield of aluminum.

experimental/theoretical (5.10g/6.22 g)*100 =

39.4 g of Al2(SO4)3 needed to go with the Mg

6.22 g of Al can be produced{and of course its the same even when starting with 39.4 g of Al2(SO4)3 }

82 % yield

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Last one……Write an overall balanced equation that represents the reaction between a solution of aluminum chloride and sodium carbonate.

5.00g * 1mol/133.33g * 1 Al2(CO3)3/2AlCl3 * 234g/mol = 4.39 g of Al2(CO3)3 precipitate can be formed

b) What would be the minimum mass of sodium carbonate we would need?5.00 g * 1mol/133.33g * 3Na2CO3/2AlCl3 * 106g/mol = 5.96 g of

Na2CO3 is required

2AlCl3 + 3Na2CO3 → Al2(CO3)3 + 6NaCla) If you had 5.00 g of aluminum chloride, (and an excess of sodium carbonate) what mass of precipitate would you expect to produce?

c) Suppose you actually did this experiment with the quantities described above, and you were able to produce 4.20 g of aluminum carbonate. Calculate the % yield of precipitate.

(exp 4.20g/theor 4.39g) * 100 = 95.7% yield